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## Comment on

Sum of Squares## I eventually used the method

I rearranged x-y = 6 in terms of y:

x = 6 + y, and then i substituted this into xy = 8

(6 + y)y = 8, then 6y + y^2 - 8 = 0 so I factored the quadratic:

(y+8)(y-2) = 0, therefore, y = -8 or 2. This is where I ran into issues:

When substituting y = 2 into xy = 8, x = 8/2 = 4. However, if I substitute y = 2 into the other equation, x - y = 6, x = 6 + (2), and then x = 8.

Why am I getting two different values for x using the same y? This is only with one of the two values for y from the quadratic.

Clearly this is wrong but I am not sure what the underlying principle of why? Could you please help?

## Everything was going fine

Everything was going fine until you got here: 6y + y² - 8 = 0 (i.e., y² + 6y - 8 = 0

When you factored the quadratic, you got: (y+8)(y-2) = 0.

This part is incorrect. If we expand (y+8)(y-2), we get: y² + 6y - 16.

Unfortunately, the expression y² + 6y - 8 cannot be factored easily (using integers), so unless we see the fast approach (as shown in the video), we'll have to apply the Quadratic Formula, which is very time consuming.

therefore, y = -8 or 2. This is where I ran into issues

## using the quadratic formula y

## Can you tell me the steps you

Can you tell me the steps you took when you applied the quadratic formula?

I ask because, when you apply the formula, you SHOULD get y = -3 + √17 and y = -3 - √17 (not y = -4 and y = -2, as you have suggested)

Cheers,

Brent

## No matter how I try, I get 20

x=6-y, therefore: y(6-y)=8

6y-y^2=8

-y^2+6y-8=0

y^2-6y+8=0

D=(-6)^2-4*8*1=36-32=4

y1=(6+2)/2=4; then x1=2

y2=(6-2)/2=2; then x2=4

either way you get: 2^2 + 4^2 = 20.

Thanks!

## Given: x - y = 6

Given: x - y = 6

So, x = y + 6 (not x = 6-y, as you have)

Cheers,

Brent

## Well, this is quite

## You're certainly not the

You're certainly not the first person to make that kind of error :-)

## Hi Brent,

In my foundations study for the GMAT, I read something about how you should not perform the Quadratic function unless the equation equals zero.

For instance, you start with (x-y)=6

-->(x-y)^2=6^2

-->x^2-2xy+y^2=36

According to what what taught in the Foundations book I read, we should not be performing this function. Rather, we would have to move 6 to the RHS of the equation which gives us x-y-6=0..However if we square this, we get a completely different formula.

I just want to know if what was taught in this book was wrong and we are able to perform a quadratic function even if the equation is not set to/equals zero.

Thanks!

## It really depends on what you

It really depends on what you're trying to accomplish.

When we solve a quadratic equation by factoring, it certainly helps to set the equation equal to zero. Here's why:

If A and B are numbers, and I tell you that AB = 0, then I can make the following conclusion with absolute certainty:

Either A = 0, B = 0 or they both = 0

This is a very strong and a very useful conclusion.

Conversely, if I tell you that AB = 12, what conclusions can you make about A and B?

Nothing.

So, when solving quadratic equations, it's very useful to set the equation equal to zero.

That said, we aren't really trying to solve an equation here.

When solving an equation, we're trying to find the value of some variable.

For example, we may want to find the value of x in the equation x² + 5x + 6 = 0

However, the question does not ask us to determine the individual values of x and y.

Instead, we're asked to find the value of x² + y², which is a different matter.

Does that help?

Cheers,

Brent

## Awesome, thank you for the